I also found that I did the last two wrong (thanx in large part by rebesque), which I have changed in my post.Originally Posted byQuick

Results 31 to 39 of 39

- Aug 25th 2006, 11:10 AM #31

- Aug 25th 2006, 01:47 PM #32

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

Actually,

$\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=\frac{\pi}{180}$. You can see that it is simpler in radians. That limit is very important.

---

Okay,

#4.

Note that,

$\displaystyle \lim_{x\to \pi/2^+}=-\infty$, yes it is true to say the limit does not exist, but I wanted you to state that is keeps getting smaller which is what this shows.

And,

$\displaystyle \lim_{x\to \pi/2^-}=\infty$, yes it is true to say the limit does not exist, but I wanted you to state that is keeps getting larger which is what this shows.

#3 is confusing.

(In fact, I did not think this was going to be difficult).

Note you**cannot**approach the limit*from the left*because, you have a number less than 1 and when you take its reciprical it is larger than 1, and when from 1 you subtract this value (which is larger) you have a negative number. But you cannot have a negative number in the radical, thus it does not exist. Thus,

$\displaystyle \lim_{x\to 1^-}\sqrt{1-\frac{1}{x}}$ does not exist and it is not even one of those infinite limits.

But,

$\displaystyle \lim_{x\to 1^+}\sqrt{1-\frac{1}{x}}=0$

Because,

$\displaystyle \left\{ \begin{array}{cc}

x&f(x)\\

1.1&.3015\\

1.01&.0995\\

1.0001&.01$

We can see the limit is zero.

Just to make you more hungry for limits, why is the limit 0? Because we say it approaches 0. But what about -1? It also approaches -1, surly. Because 0 is less than -1. In fact, 0 is the smallest number which it approaches and does not exceede. This fabolous result is called*Weierstrauss-Bolzano Theorem*.

So, I hope you understand limits. What now? I can lecture about,

1)Conics

2)Mathematical Induction

3)More limits (this times we learn rule to compute limits).

...?

- Aug 25th 2006, 01:51 PM #33

- Aug 25th 2006, 01:54 PM #34

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

Originally Posted by**Quick**

(But not now).

- Aug 26th 2006, 05:40 PM #35

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

In Number Theory (a truly elegant topic in pure mathematics that studies how numbers and properties about them) a very useful principle for building fundamental results is called

*well-ordering principle*.

**Well-Ordering Principle**: Any non-empty set of positive integers contains a minimal element.

This intuitevly obvious principle can be shown to be true by the Axiom of Choice (sadly I do not know the details).

This Principle can be used in several very important concepts involving positive integers. One is called*induction*.

**Mathematical Induction**: If $\displaystyle S$ is a set of some positive integers and:

1) $\displaystyle 1\in S$

2)If $\displaystyle k\in S$ then $\displaystyle (k+1)\in S$.

Then,

$\displaystyle \mathbb{Z}^+\subseteq S$ meaning the positive integers are contained in this set.

**Proof**: Let $\displaystyle S'$ be the complement of S in the positive integers. Meaning let $\displaystyle S'$ be the set of all integers that are not in S. If we can show that S' is empty then we have shown that S contains all the positive integers.

We will procede by contradiction. Assume, that S' is non-empty. But the Well-Ordering Principle assures us there is a minimal positive integer in S' which will be $\displaystyle a$. Since $\displaystyle 1\in S$ (by the hypothesis) we conclude that $\displaystyle 1\not \in S'$ therefore,

$\displaystyle a>1$. Therefore, $\displaystyle 0<a-1<a$ but $\displaystyle (a-1)\not \in S'$ (because it is smaller) thus, $\displaystyle a-1\in S$ but then, $\displaystyle (a-1)+1=a\in S$ which is a contradiction. Thus, $\displaystyle S'=\{ \}$.

(Basically what the theorem is saying if k is in S then then number after that. But since the number after that is in S then the number after that... Thus all the numbers are in S).

Now how can we use this useful tool? For example, usually textbooks use the following elementray example showing its power.

Prove that,

$\displaystyle 1+2+...+n=\frac{n(n+1)}{2}$.

Let S be the set of all positive integers such that the above

equation is true.

Notice that,

$\displaystyle 1\in S$ because,

$\displaystyle 1=\frac{1(2)}{2}=1$.

Next let us say that $\displaystyle k\in S$ we will show that $\displaystyle (k+1)\in S$. Since k is in S thus,

$\displaystyle 1+2+...+k=\frac{k(k+1)}{2}$

Now, add (k+1) to both sides,

$\displaystyle 1+2+...+k+(k+1)=\frac{k(k+1)}{2}+(k+1)$

The right hand side fraction can be simplified thus,

$\displaystyle 1+2+...+k+(k+1)=\frac{(k+1)(k+2)}{2}$.

Ah! Thus $\displaystyle (k+1)\in S$. Therefore by the theorem S contains*all*the integers.

Conclusion: All positive integers are in S.

Since we defined S as the set that is composed of positive integers that make that true we see that all positive integers are in S thus it is true for all integers.

- Aug 28th 2006, 02:10 PM #36

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

You might have thought maybe sometimes we can substitute the limiting value into the function to gets is limit. For example,

$\displaystyle \lim_{x\to 1}x=1$.

Here is an important definition.

**Definition**: A function $\displaystyle f(x)$ is*countinous*at $\displaystyle x=c$ means that,

$\displaystyle \lim_{x\to c}f(x)=f(c)$

The reason why we chose such a term is because informally and visually whenever you can substitute a value for the limit you have a curve with no holes in it. Meaning it is countinous, no breaks, no jumps, no tears. It is difficult to explain to you why we can think of a continous curve such as the one that has the above property. Try to read about on the internet they should have some visual. But the important think this we have a mathematical definition to what countinous means.

Without proof we present the following useful tools for a mathematician.

**Theorem**: If $\displaystyle f(x),g(x)$ is continous at $\displaystyle x=c$ then, so is

$\displaystyle f(x)+g(x)$----> Their sum

$\displaystyle f(x)-g(x)$----->Their difference

$\displaystyle f(x)g(x)$-----> Their product

$\displaystyle f(x)/g(x), g(c)\not = 0$---->Their quoteint

$\displaystyle f(g(x))$----->Their composition.

Basically, limits can be divided into two categories. Limits such as those when you can substitute (meaning the number you are substituting is in the domain) and those that do not. The first case is extremely easy, simply substitute. The second case is where all the question on the forum are, that is where you need to work on the limit.

To help explain my previous paragraph I will explain what an elementary function is.

**Definition**: An*elementary*function is such a function composed as a sum, difference, product, quotient and composition of the: ploynomial function (contant, linear, quadradic, cubic,...), exponential, logarithmic, trigonometric and inverse trigonometric.

**Example**: The following function is elementary.

$\displaystyle \sin(x+2x+2^x-\sqrt{\cos x})$.

You can see it is composed of the "basic" functions mentioned above.

**Example**: The following function is not elementary.

$\displaystyle |x|$

The purpose of this is to show that whenever an elementary function is evaluated at its domain you can just substitute. If you every graphed the basic functions mentioned above you saw they were all continous. And since we said that combining continous function in any way produces a continous function. Therefore any elementary function is continous at a point where it is defined because by definition an elementary function is some combination of these basic continous functions.

**Example**$\displaystyle \lim_{x\to 0}\sin(x+2x+2^x-\sqrt{\cos x})$

Simple. Check whether you can substitute 0. Meaning whether or not 0 is in the domain of the function. Of course it is. So the limit is,

$\displaystyle \sin (0+0+1-1)=0$

When it is not, like here,

$\displaystyle \lim_{x\to 0}\frac{x}{x}=1$

You cannot because you cannot 0/0.

However,

$\displaystyle \lim_{x\to 1}\frac{x}{x}=\frac{1}{1}=1$

You can.

---

Now we turn to the next discussion what to do when you cannot substitute? Most of the limits are expressed in fraction. If by substituting you get,

$\displaystyle \frac{n}{0}, n\not =0$ then your limit is an infinite limit, like the ones discussed before. If the value of "n" is positive you get a postive infinite limit otherwise a negative infinite limit. Such as,

$\displaystyle \lim_{x\to 0}\frac{1}{x}$

If you substitute you get, $\displaystyle 1/0$ so the limit is $\displaystyle +\infty$. (Again it does not mean that the limit exist!). The entire problem happens when you n=0. That is for another discussion.

- Sep 2nd 2006, 06:36 PM #37

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

Now we discuss the complicated question. What do you do when the limit of the numerator and denominator is zero, i.e. it gives the form 0/0.

**Defintion**: $\displaystyle (a,b)=a<x<b$ is called the*open-interval*. $\displaystyle [a,b]=a\leq x\leq b$ is called*closed-interval*.

Here is a useful theorem for evaluating limits.

**Theorem**: If two functions agree at an open interval containing $\displaystyle c$ except at possibly at $\displaystyle c$ itself. Then there limits are equal if they exist.

**Proof**: We will not state a formal prove of this but we can intuitevly imagine this. The entire concept of limits is based on the fact that we approach a point closer and closer. Thus, all what matters is that the two functions are identitical close to $\displaystyle c$ and what is far away makes no difference. Also, it makes no difference if the two functions agree (meaning are identical) at $\displaystyle c$ because you are approaching it. Finally, the reason why we say an*open*interval is because we emphasize that they agree from the right (close to it) and from the left (close to it) because open interval includes $\displaystyle c$ in between. If we said a*closed*interval we would not know that because it can include $\displaystyle c$ as an endpoint.

**Example**: Find the limit of $\displaystyle \lim_{x\to 0}\frac{x}{x}$. You might be tempted to cancel the x's and say $\displaystyle f(x)=1$ and then say $\displaystyle \lim_{x\to 0}1=1$ (because it is a constant function and we learned before constant functions are countinous). There is a problem with this approach, $\displaystyle 1\not = x/x$! Because the function $\displaystyle f(x)=1$ has domain all real numbers, but the function $\displaystyle g(x)=x/x$ has domain all real numbers expcept for zero. Since the two functions are not equal then we cannot conclude that. However, we can employ the previous theorem. Chose any open interval say $\displaystyle (-1,1)$ (in fact any works) that contains $\displaystyle c=0$. Note the functions*agree*on it except at possibly $\displaystyle c=0$ because both outputs give 1. Therefore, the limit of $\displaystyle x/x$ behaves just like $\displaystyle f(x)=1$ and we conclude that the limit is 1.

**Example**: Find the limits of,

$\displaystyle \lim_{x\to 1}\frac{x^2-1}{x-1}$. First we factor.

$\displaystyle \frac{(x-1)(x+1)}{(x-1)}=x+1,x\not = 1$

And we use the same argument as before, that we say that though the two function are not the same (because it is not defined for 1) but they are the same except for the limit point so the limits is the same if they exist,

$\displaystyle \lim_{x\to 1}x+1=1+1=2$ and of course this evaluation as true as explained before about continous functions.

**Example**: Find,

$\displaystyle \lim_{x\to 4}\frac{x-4}{\sqrt{x}-2}$.

Listen how to do this example then you can start answering some of the limit questions on the forums. The basic procedure they teach in college is to*rationalize*.

Thus,

$\displaystyle \frac{x-4}{\sqrt{x}-2}\cdot \frac{\sqrt{x}+2}{\sqrt{x}+2}$

This will elimanite the zero denominator and thus you can start using substitution because you will turn it into a countinous function at that point. Of couse, you use the same reasoning to why this works as in the last two examples.

Another, useful theorem is given below. My favorite.

**Squeeze Theorem**: If a function is*squeezed*between two functions (meaning it is in between them in value) on some open interval containing $\displaystyle c$ and the lower bound and upper bound both exist at that limit with equal values then so does the function with the same value.

Sounds complicated, right? It is not a very efficient method because you need to think of two simple functions that can squeeze your function to evaluate the limit. It is an extremely useful theoretical tool (meaning to prove theorems).

**Example**: Find,

$\displaystyle \lim_{x\to 0}x\sin (1/x)$ we cannot substitute because it is not a countinous function. However, for any open interval containing $\displaystyle c=0$ we have,

$\displaystyle -1 \leq \sin (1/x)\leq 1$. Thus,

$\displaystyle -x\leq x\sin (1/x)\leq x$ (if x is positve the sign does not change if negative the sign changes but the inequality is still preserved). We note that our function for which we are trying to find the limit is squeezed between the lower bound $\displaystyle -x$ and upper bound $\displaystyle x$ both of which have the same limit at zero, i.e. 0. Thus we conclude that the answer to our problem is too zero.

Here are two important trigonometric limits I will prove (informally but still nice) later involving the squeeze theorem.

$\displaystyle \frac{\sin x}{x}=1,x\to 0$

$\displaystyle \frac{1-\cos x}{x}=0,x\to 0$

- May 26th 2007, 06:15 PM #38

- May 26th 2007, 06:20 PM #39

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10